Đáp án: M=1
Giải thích các bước giải:
$\begin{array}{l}
3{x^2} + 3{y^2} + 4xy + 2x - 2y + 2 = 0\\
\Rightarrow 2{x^2} + 2{y^2} + 4xy + {x^2} + {y^2} + 2x - 2y + 2 = 0\\
\Rightarrow 2\left( {{x^2} + {y^2} + 2xy} \right) + \left( {{x^2} + 2x + 1} \right) + \left( {{y^2} - 2y + 1} \right) = 0\\
\Rightarrow 2{\left( {x + y} \right)^2} + {\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
2{\left( {x + y} \right)^2} \ge 0\forall x,y\\
{\left( {x + 1} \right)^2} \ge 0\forall x\\
{\left( {y - 1} \right)^2} \ge 0\forall y
\end{array} \right.\\
Nên:2{\left( {x + y} \right)^2} + {\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} = 0\\
\Rightarrow \left\{ \begin{array}{l}
x + y = 0\\
x + 1 = 0\\
y - 1 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - y\\
x = - 1\\
y = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - 1\\
y = 1
\end{array} \right.\\
M = {\left( {x + y} \right)^{2010}} + {\left( {x + 2} \right)^{2011}} + {\left( {y - 1} \right)^{2012}}\\
= {0^{2010}} + {\left( { - 1 + 2} \right)^{2011}} + {0^{2012}}\\
= {1^{2011}}\\
= 1
\end{array}$
